Question

Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.9 m/s. Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?

Answer 1

A)
we can treat basket ball as hollow sphere.

Apply conservtion of energy

final kinetic energy = initial potential energy

(1/2)*m*v^2 + (1/2)*I*w^2 = m*g*h

(1/2)*m*v^2 + (1/2)*(2/3)*m*r^2*w^2 = m*g*h

(1/2)*m*v^2 + (1/3)*m*(r*w)^2 = m*g*h

(1/2)*m*v^2 + (1/3)*m*v^2 = m*g*h

0.833*m*v^2 = m*g*h

==> h = 0.833*v^2/g

= 0.833*6.9^2/9.8

= 4.05 m <<<<<<<<<<-----------------------Answer

b) we can treat the can of frozen juice with cyllinder.

final kinetic energy = initial potential energy

(1/2)*m*v^2 + (1/2)*I*w^2 = m*g*h

(1/2)*m*v^2 + (1/2)*(1/2)*m*r^2*w^2 = m*g*h

(1/2)*m*v^2 + (1/4)*m*(r*w)^2 = m*g*h

(1/2)*m*v^2 + (1/4)*m*v^2 = m*g*h

0.75*m*v^2 = m*g*h

v = sqrt(g*h/0.75)

= sqrt(9.8*4.05/0.75)

= 7.3 m/s <<<<<<<<<<<<----------------------Answer